12.6: Adding and Subtracting Rational Expressions
Learning Objectives
- Add and subtract rational expressions with the same denominator.
- Find the least common denominator of rational expressions.
- Add and subtract rational expressions with different denominators.
- Solve real-world problems involving addition and subtraction of rational expressions.
Introduction
Like fractions, rational expressions represent a portion of a quantity. Remember that when we add or subtract fractions we must first make sure that they have the same denominator. Once the fractions have the same denominator, we combine the different portions by adding or subtracting the numerators and writing that answer over the common denominator.
Add and Subtract Rational Expressions with the Same Denominator
Fractions with common denominators combine in the following manner:
\begin{align*}\frac{a}{c}+\frac{b}{c} = \frac{a+b}{c} \qquad \text{and} \qquad \frac{a}{c} - \frac{b}{c}=\frac{a-b}{c}\end{align*}
Example 1
Simplify.
a) \begin{align*}\frac{8}{7} - \frac{2}{7} + \frac{4}{7}\end{align*}
b) \begin{align*}\frac{4x^2-3}{x+5} + \frac{2x^2-1}{x+5}\end{align*}
c) \begin{align*}\frac{x^2-2x+1}{2x+3} - \frac{3x^2-3x+5}{2x+3}\end{align*}
Solution
a) Since the denominators are the same we combine the numerators:
\begin{align*}\frac{8}{7} - \frac{2}{7} + \frac{4}{7} = \frac{8-2+4}{7} = \frac{10}{7}\end{align*}
b) \begin{align*}\text{Since the denominators are the same we combine the numerators:} \qquad \frac{4x^2-3+2x^2-1}{x+5}\!
\\
\text{Simplify by collecting like terms:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{6x^2-4}{x+5}\end{align*}
c) Since the denominators are the same we combine the numerators. Make sure the subtraction sign is distributed to all terms in the second expression:
\begin{align*}\frac{x^2-2x+1-(3x^2-3x+5)}{2x+3} = \frac{x^2-2x+1-3x^2+3x-5}{2x+3}= \frac{-2x^2+x-4}{2x+3}\end{align*}
Find the Least Common Denominator of Rational Expressions
To add and subtract fractions with different denominators, we must first rewrite all fractions so that they have the same denominator. In general, we want to find the least common denominator. To find the least common denominator, we find the least common multiple (LCM) of the expressions in the denominators of the different fractions. Remember that the least common multiple of two or more integers is the least positive integer that has all of those integers as factors.
The procedure for finding the lowest common multiple of polynomials is similar. We rewrite each polynomial in factored form and we form the LCM by taking each factor to the highest power it appears in any of the separate expressions.
Example 2
Find the LCM of \begin{align*}48x^2y\end{align*}
Solution
First rewrite the integers in their prime factorization.
\begin{align*}48 & = 2^4 \cdot 3\\
60 & = 2^2 \cdot 3 \cdot 5\end{align*}
The two expressions can be written as:
\begin{align*}& 48x^2y=2^4 \cdot 3 \cdot x^2 \cdot y\\
& 60xy^3z=2^2 \cdot 3 \cdot 5 \cdot x \cdot y^3 \cdot z\end{align*}
To find the LCM, take the highest power of each factor that appears in either expression.
\begin{align*}\text{LCM} = 2^4 \cdot 3 \cdot 5 \cdot x^2 \cdot y^3 \cdot z = 240x^2y^3z\end{align*}
Example 3
Find the LCM of \begin{align*}2x^2+8x+8\end{align*}
Solution
Factor the polynomials completely:
\begin{align*}2x^2+8x+8 & = 2(x^2+4x+4)\\
& = 2(x+2)^2\end{align*}
\begin{align*}x^3-4x^2-12x & = x(x^2-4x-12)\\
& = x(x+2)(x-6)\end{align*}
To find the LCM, take the highest power of each factor that appears in either expression.
\begin{align*}\text{LCM} = 2x(x+2)^2 (x-6)\end{align*}
It’s customary to leave the LCM in factored form, because this form is useful in simplifying rational expressions and finding any excluded values.
Example 4
Find the LCM of \begin{align*}x^2-25\end{align*}
Solution
Factor the polynomials completely:
\begin{align*}x^2-25 & = (x-5)(x+5)\\
x^2+3x+2 & = (x+1)(x+2)\end{align*}
Since the two expressions have no common factors, the LCM is just the product of the two expressions.
\begin{align*}\text{LCM} = (x-5)(x+5)(x+1)(x+2)\end{align*}
Add and Subtract Rational Expressions with Different Denominators
Now we’re ready to add and subtract rational expressions. We use the following procedure.
- Find the least common denominator (LCD) of the fractions.
- Express each fraction as an equivalent fraction with the LCD as the denominator.
- Add or subtract and simplify the result.
Example 5
Perform the following operation and simplify: \begin{align*}\frac{2}{x+2} - \frac{3}{2x-5}\end{align*}
Solution
The denominators can’t be factored any further, so the LCD is just the product of the separate denominators: \begin{align*}(x+2)(2x-5)\end{align*}. That means the first fraction needs to be multiplied by the factor \begin{align*}(2x-5)\end{align*} and the second fraction needs to be multiplied by the factor \begin{align*}(x+2)\end{align*}:
\begin{align*}\frac{2}{x+2} \cdot \frac{(2x-5)}{(2x-5)} - \frac{3}{2x-5} \cdot \frac{(x+2)}{(x+2)}\end{align*}
\begin{align*}\text{Combine the numerators and simplify:} \qquad \qquad \frac{2(2x-5)-3(x+2)}{(x+2)(2x-5)} = \frac{4x-10-3x-6}{(x+2)(2x-5)}\!\\ \\ \text{Combine like terms in the numerator:} \qquad \qquad \frac{x-16}{(x+2)(2x-5)} \quad \mathbf{Answer}\end{align*}
Example 6
Perform the following operation and simplify: \begin{align*}\frac{4x}{x-5}-\frac{3x}{5-x}\end{align*}.
Solution
Notice that the denominators are almost the same; they just differ by a factor of -1.
\begin{align*}\text{Factor out -1 from the second denominator:} \qquad \qquad \qquad \qquad \qquad \qquad \frac{4x}{x-5} - \frac{3x}{-(x-5)}\!\\ \\ \text{The two negative signs in the second fraction cancel:} \qquad \qquad \qquad \qquad \frac{4x}{x-5}+\frac{3x}{(x-5)}\!\\ \\ \text{Since the denominators are the same we combine the numerators:} \ \qquad \frac{7x}{x-5} \quad \mathbf{Answer}\end{align*}
Example 7
Perform the following operation and simplify: \begin{align*}\frac{2x-1}{x^2-6x+9}-\frac{3x+4}{x^2-9}\end{align*}.
Solution
\begin{align*}\text{We factor the denominators:} \qquad \qquad \frac{2x-1}{(x-3)^2}-\frac{3x+4}{(x+3)(x-3)}\end{align*}
The LCD is the product of all the different factors, each taken to the highest power they have in either denominator: \begin{align*}(x-3)^2(x+3)\end{align*}.
The first fraction needs to be multiplied by a factor of \begin{align*}(x + 3)\end{align*} and the second fraction needs to be multiplied by a factor of \begin{align*}(x - 3)\end{align*}:
\begin{align*}\frac{2x-1}{(x-3)^2} \cdot \frac{(x+3)}{(x+3)}-\frac{3x+4}{(x+3)(x-3)} \cdot \frac{(x-3)}{(x-3)}\end{align*}
\begin{align*}\text{Combine the numerators by subtracting:} \qquad \qquad \qquad \qquad \frac{(2x-1)(x+3)-(3x+4)(x-3)}{(x-3)^2(x+3)}\!\\ \\ \text{Eliminate parentheses in the numerator:} \qquad \qquad \qquad \qquad \frac{2x^2+5x-3-(3x^2-5x-12)}{(x-3)^2(x+3)}\!\\ \\ \text{Distribute the negative sign:} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \frac{2x^2+5x-3-3x^2+5x+12}{(x-3)^2(x+3)}\!\\ \\ \text{Combine like terms in the numerator:} \qquad \qquad \qquad \qquad \quad \ \frac{-x^2+10x+9}{(x-3)^2(x+3)} \quad \mathbf{Answer}\end{align*}
For more examples of how to add and subtract rational expressions, watch the video at
.
Solve Real-World Problems by Adding and Subtracting Rational Expressions
Example 8
In an electrical circuit with two resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance: \begin{align*}\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}\end{align*}. Find an expression for the total resistance, \begin{align*}R_{tot}\end{align*}.
Solution
\begin{align*}\text{Let's simplify the expression} \ \frac{1}{R_1}+\frac{1}{R_2}.\!\\ \!\\ \text{The lowest common denominator is} \ R_1R_2, \ \text{so we multiply the first fraction by} \ \frac{R_2}{R_2} \ \text{and the}\!\\ \!\\ \text{second fraction by} \ \frac{R_1}{R_1}: \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{R_2}{R_2} \cdot \frac{1}{R_1} + \frac{R_1}{R_1} \cdot \frac{1}{R_2}\!\\ \!\\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \frac{R_2 + R_1}{R_1R_2}\!\\ \!\\ \text{The total resistance is the reciprocal of this expression:} \qquad R_{tot}=\frac{R_1R_2}{R_1+R_2} \quad \mathbf{Answer}\end{align*}
Example 9
The sum of a number and its reciprocal is \begin{align*}\frac{53}{14}\end{align*}. Find the numbers.
Solution
Define variables:
Let \begin{align*}x\end{align*} be the number; then its reciprocal is \begin{align*}\frac{1}{x}\end{align*}.
Set up an equation:
The equation that describes the relationship between the numbers is \begin{align*}x+\frac{1}{x}=\frac{53}{14}\end{align*}
Solve the equation:
\begin{align*}\text{Find the lowest common denominator:} \ \qquad \text{LCM} = 14x\!\\ \!\\ \text{Multiply all terms by} \ 14x: \qquad \qquad \qquad \quad \ 14x \cdot x + 14x \cdot \frac{1}{x}=14x \cdot \frac{53}{14}\end{align*}
(Notice that we’re multiplying the terms by \begin{align*}14x\end{align*} instead of by \begin{align*}\frac{14x}{14x}\end{align*}. We can do this because we’re multiplying both sides of the equation by the same thing, so we don’t have to keep the actual values of the terms the same. We could also multiply by \begin{align*}\frac{14x}{14x}\end{align*}, but then the denominators would just cancel out a couple of steps later.)
\begin{align*}\text{Cancel common factors in each term:} \qquad \qquad \ 14x \cdot x + 14x \cdot \frac{1}{x} = 14x \cdot \frac{53}{14}\!\\ \!\\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 14x^2 + 14 = 53x\!\\ \!\\ \text{Write all terms on one side of the equation:} \qquad 14x^2 - 53x + 14 = 0\!\\ \!\\ \text{Factor:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (7x-2)(2x-7) = 0\!\\ \!\\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x=\frac{2}{7} \ \text{and} \ x=\frac{7}{2}\end{align*}
Notice there are two answers for \begin{align*}x\end{align*}, but they are really parts of the same solution. One answer represents the number and the other answer represents its reciprocal.
Check:
\begin{align*}\frac{2}{7}+\frac{7}{2}=\frac{4+49}{14}=\frac{53}{14}\end{align*}. The answer checks out.
Work problems are problems where two people or two machines work together to complete a job. Work problems often contain rational expressions. Typically we set up such problems by looking at the part of the task completed by each person or machine. The completed task is the sum of the parts of the tasks completed by each individual or each machine.
To determine the part of the task completed by each person or machine we use the following fact:
\begin{align*}\text{Part of the task completed} = \text{rate of work} \times \text{time spent on the task}\end{align*}
It’s usually useful to set up a table where we can list all the known and unknown variables for each person or machine and then combine the parts of the task completed by each person or machine at the end.
Example 10
Mary can paint a house by herself in 12 hours. John can paint a house by himself in 16 hours. How long would it take them to paint the house if they worked together?
Solution
Define variables:
Let \begin{align*}t =\end{align*} the time it takes Mary and John to paint the house together.
Construct a table:
Since Mary takes 12 hours to paint the house by herself, in one hour she paints \begin{align*}\frac{1}{12}\end{align*} of the house.
Since John takes 16 hours to pain the house by himself, in one hour he paints \begin{align*}\frac{1}{16}\end{align*} of the house.
Mary and John work together for \begin{align*}t\end{align*} hours to paint the house together. Using
\begin{align*}Part \ of \ the \ task \ completed = rate \ of \ work \cdot time \ spent \ on \ the \ task\end{align*}
we can write that Mary completed \begin{align*}\frac{t}{12}\end{align*} of the house and John completed \begin{align*}\frac{t}{16}\end{align*} of the house in this time.
This information is nicely summarized in the table below:
Painter | Rate of work (per hour) | Time worked | Part of task |
---|---|---|---|
Mary | \begin{align*}\frac{1}{12}\end{align*} | \begin{align*}t\end{align*} | \begin{align*}\frac{t}{12}\end{align*} |
John | \begin{align*}\frac{1}{16}\end{align*} | \begin{align*}t\end{align*} | \begin{align*}\frac{t}{16}\end{align*} |
Set up an equation:
In \begin{align*}t\end{align*} hours, Mary painted \begin{align*}\frac{t}{12}\end{align*} of the house and John painted \begin{align*}\frac{t}{16}\end{align*} of the house, and together they painted 1 whole house. So our equation is \begin{align*}\frac{t}{12}+\frac{t}{16}=1\end{align*}.
Solve the equation:
\begin{align*}\text{Find the lowest common denominator:} \qquad \qquad \quad \qquad \text{LCM} = 48\!\\ \\ \text{Multiply all terms in the equation by the LCM:} \qquad \ \ 48 \cdot \frac{t}{12}+48 \cdot \frac{t}{16}=48 \cdot 1\!\\ \\ \text{Cancel common factors in each term:} \qquad \qquad \qquad \quad \ 4 \cdot \frac{t}{1}+3 \cdot \frac{t}{1}=48 \cdot 1\!\\ \\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 4t+3t=48\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 7t=48 \Rightarrow t=\frac{48}{7}=6.86 \ hours\end{align*}
Check: The answer is reasonable. We’d expect the job to take more than half the time Mary would take by herself but less than half the time John would take, since Mary works faster than John.
Example 11
Suzie and Mike take two hours to mow a lawn when they work together. It takes Suzie 3.5 hours to mow the same lawn if she works by herself. How long would it take Mike to mow the same lawn if he worked alone?
Solution
Define variables:
Let \begin{align*}t =\end{align*} the time it takes Mike to mow the lawn by himself.
Construct a table:
Painter | Rate of work (per hour) | Time worked | Part of Task |
---|---|---|---|
Suzie | \begin{align*}\frac{1}{3.5}=\frac{2}{7}\end{align*} | 2 | \begin{align*}\frac{4}{7}\end{align*} |
Mike | \begin{align*}\frac{1}{t}\end{align*} | 2 | \begin{align*}\frac{2}{t}\end{align*} |
Set up an equation:
Since Suzie completed \begin{align*}\frac{4}{7}\end{align*} of the lawn and Mike completed \begin{align*}\frac{2}{t}\end{align*} of the lawn and together they mowed the lawn in 2 hours, we can write the equation: \begin{align*}\frac{4}{7}+\frac{2}{t}=1\end{align*}
Solve the equation:
\begin{align*}\text{Find the lowest common denominator:} \qquad \qquad \quad \qquad \text{LCM} = 7t\!\\ \\ \text{Multiply all terms in the equation by the LCM:} \qquad \ \ 7t \cdot \frac{4}{7}+7t \cdot \frac{2}{t}=7t \cdot 1\!\\ \\ \text{Cancel common factors in each term:} \qquad \qquad \qquad \quad \ t \cdot \frac{4}{1}+7 \cdot \frac{2}{1}=7t \cdot 1\!\\ \\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 4t+14=7t\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 3t=14 \Rightarrow t=\frac{14}{3}=4 \frac{2}{3} \ hours\end{align*}
Check: The answer is reasonable. We’d expect Mike to work slower than Suzie, because working by herself it takes her less than twice the time it takes them to work together.
Review Questions
Perform the indicated operation and simplify. Leave the denominator in factored form.
- \begin{align*}\frac{5}{24}-\frac{7}{24}\end{align*}
- \begin{align*}\frac{10}{21}+\frac{9}{35}\end{align*}
- \begin{align*}\frac{5}{2x+3}+\frac{3}{2x+3}\end{align*}
- \begin{align*}\frac{3x-1}{x+9}-\frac{4x+3}{x+9}\end{align*}
- \begin{align*}\frac{4x+7}{2x^2}-\frac{3x-4}{2x^2}\end{align*}
- \begin{align*}\frac{x^2}{x+5}-\frac{25}{x+5}\end{align*}
- \begin{align*}\frac{2x}{x-4}+\frac{x}{4-x}\end{align*}
- \begin{align*}\frac{10}{3x-1}-\frac{7}{1-3x}\end{align*}
- \begin{align*}\frac{5}{2x+3}-3\end{align*}
- \begin{align*}\frac{5x+1}{x+4}+2\end{align*}
- \begin{align*}\frac{1}{x}+\frac{2}{3x}\end{align*}
- \begin{align*}\frac{4}{5x^2}-\frac{2}{7x^3}\end{align*}
- \begin{align*}\frac{4x}{x+1}-\frac{2}{2(x+1)}\end{align*}
- \begin{align*}\frac{10}{x+5}+\frac{2}{x+2}\end{align*}
- \begin{align*}\frac{2x}{x-3}-\frac{3x}{x+4}\end{align*}
- \begin{align*}\frac{4x-3}{2x+1}+\frac{x+2}{x-9}\end{align*}
- \begin{align*}\frac{x^2}{x+4}-\frac{3x^2}{4x-1}\end{align*}
- \begin{align*}\frac{2}{5x+2}-\frac{x+1}{x^2}\end{align*}
- \begin{align*}\frac{x+4}{2x}+\frac{2}{9x}\end{align*}
- \begin{align*}\frac{5x+3}{x^2+x}+\frac{2x+1}{x}\end{align*}
- \begin{align*}\frac{4}{(x+1)(x-1)}-\frac{5}{(x+1)(x+2)}\end{align*}
- \begin{align*}\frac{2x}{(x+2)(3x-4)}+\frac{7x}{(3x-4)^2}\end{align*}
- \begin{align*}\frac{3x+5}{x(x-1)}-\frac{9x-1}{(x-1)^2}\end{align*}
- \begin{align*}\frac{1}{(x-2)(x-3)}+\frac{4}{(2x+5)(x-6)}\end{align*}
- \begin{align*}\frac{3x-2}{x-2}+\frac{1}{x^2-4x+4}\end{align*}
- \begin{align*}\frac{-x^3}{x^2-7x+6}+x-4\end{align*}
- \begin{align*}\frac{2x}{x^2+10x+25}-\frac{3x}{2x^2+7x-15}\end{align*}
- \begin{align*}\frac{1}{x^2-9}+\frac{2}{x^2+5x+6}\end{align*}
- \begin{align*}\frac{-x+4}{2x^2-x-15}+\frac{x}{4x^2+8x-5}\end{align*}
- \begin{align*}\frac{4}{9x^2-49}-\frac{1}{3x^2+5x-28}\end{align*}
- One number is 5 less than another. The sum of their reciprocals is \begin{align*}\frac{13}{36}\end{align*}. Find the two numbers.
- One number is 8 times more than another. The difference in their reciprocals is \begin{align*}\frac{21}{20}\end{align*}. Find the two numbers.
- A pipe can fill a tank full of Kool-Aid in 4 hours and another pipe can empty the tank in 8 hours. If the valves to both pipes are open, how long will it take to fill the tank?
- Stefan and Misha have a lot full of cars to wash. Stefan could wash the cars by himself in 6 hours and Misha could wash the cars by himself in 5 hours. Stefan starts washing the cars by himself, but he has to leave after 2.5 hours. Misha continues the task by himself. How long does it take Misha to finish washing the cars?
- Amanda and her sister Chyna are shoveling snow to clear their driveway. Amanda can clear the snow by herself in 3 hours and Chyna can clear the snow by herself in 4 hours. After Amanda has been working by herself for one hour, Chyna joins her and they finish the job together. How long does it take to clear the snow from the driveway?
- At a soda bottling plant one bottling machine can fulfill the daily quota in 10 hours and a second machine can fill the daily quota in 14 hours. The two machines start working together, but after four hours the slower machine breaks and the faster machine has to complete the job by itself. How many more hours does the fast machine take to finish the job?
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